HDU3255 Farming（线段树+扫描线）

Farming

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1614    Accepted Submission(s): 479

Problem Description

You have a big farm, and you want to grow vegetables in it. You're too lazy to seed the seeds yourself, so you've hired n people to do the job for you.
Each person works in a rectangular piece of land, seeding one seed in one unit square. The working areas of different people may overlap, so one unit square can be seeded several times. However, due to limited space, different seeds in one square fight each other -- finally, the most powerful seed wins. If there are several "most powerful" seeds, one of them win (it does not matter which one wins).

There are m kinds of seeds. Different seeds grow up into different vegetables and sells for different prices.
As a rule, more powerful seeds always grow up into more expensive vegetables.
Your task is to calculate how much money will you get, by selling all the vegetables in the whole farm.

 

Input

The first line contains a single integer T (T <= 10), the number of test cases.
Each case begins with two integers n, m (1 <= n <= 30000, 1 <= m <= 3).
The next line contains m distinct positive integers p _i (1 <= p _i <= 100), the prices of each kind of vegetable.
The vegetables (and their corresponding seeds) are numbered 1 to m in the order they appear in the input.
Each of the following n lines contains five integers x1, y1, x2, y2, s, indicating a working seeded a rectangular area with lower-left corner (x1,y1), upper-right corner (x2,y2), with the s-th kind of seed.
All of x1, y1, x2, y2 will be no larger than 10 ⁶ in their absolute values.

 

Output

For each test case, print the case number and your final income.

 

Sample Input

  

   2
1 1
25
0 0 10 10 1
2 2
5 2
0 0 2 1 1
1 0 3 2 2
  

 

Sample Output

  

   Case 1: 2500
Case 2: 16
  

 

Source

2009 Asia Regional Ningbo Online

题意：给出n个矩形，每块矩形种不同价值的种子，矩形会出现相交情况，这时候价值高的种子会把价值低的种子淘汰，问最后得到全部价值。

思路：

如果线段树+扫描线可以求覆盖面积，那么这题就是求体积并，试想把种子的价值当作高，那么按高度升序排序后，分段求体积，选出高度不小于h[i]的矩形插入线段树求当前底面积，每次就相当于求一边覆盖面积（其实就是当前底面积，那么久相当同一块矩形里，最高的高（价值最大的种子）会被当作高去求体积），求得的覆盖面积*（h[i+1]-h[i]）。那么把每次求得的分段体积加起来就是总的体积了。

线段树+扫描线求覆盖面积： 点击打开链接

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 35000
#define ll long long int
ll sum[maxn<<2];
int mark[maxn<<2], val[5], Hash[maxn<<1];
struct seg{
    int l, r, h, c, val;
    seg(){}
    seg(int x1, int x2, int y, int v, int d):l(x1),r(x2),h(y),val(v),c(d){}
    bool operator <(const seg &a)const{
        return h<a.h;
    }
}s[maxn<<1], tmp[maxn<<1];
int Search(int key, int a[], int right){
    int l = 0, r = right-1;
    while(l <= r){
        int mid = (l+r)>>1;
        if(a[mid] == key) return mid;
        if(a[mid] > key) r = mid-1;
        else l = mid+1;
    }
    return -1;
}
void upfather(int n, int left, int right){
    if(mark[n]) sum[n] = Hash[right+1] - Hash[left];
    else if(left == right) sum[n] = 0;
    else sum[n] = sum[n<<1] + sum[n<<1|1];
}
void update(int L, int R, int rt, int d, int left, int right){
    if(L<=left&&right<=R){
        mark[rt]+=d;
        upfather(rt, left, right);
        return;
    }
    int mid = (left+right)>>1;
    if(L<=mid) update(L, R, rt<<1, d, left, mid);
    if(mid<R) update(L, R, rt<<1|1, d, mid+1, right);
    upfather(rt, left, right);
}
int main()
{
    int T, k, i, j, n, m, x1, x2, y1, y2, Case, v, p;
    scanf("%d", &T);
    Case = 1;
    while(T--){
        scanf("%d %d", &n, &m);
        for(i = 1;i <= m;i++) scanf("%d", &val[i]);
        k = 0;
        for(i = 0;i < n;i++){
            scanf("%d %d %d %d %d", &x1, &y1, &x2, &y2, &v);
            Hash[k]= x1;
            s[k++] = seg(x1, x2, y1, val[v], 1);
            Hash[k]= x2;
            s[k++] = seg(x1, x2, y2, val[v], -1);
        }
        val[0] = 0;
        sort(Hash, Hash+k);
        sort(s, s+k);
        sort(val, val+m+1);
        p = 1;
        for(i = 1;i < k;i++)
            if(Hash[i] != Hash[i-1])
                Hash[p++] = Hash[i];
        ll ans = 0;
        for(i = 1;i <= m;i++){
            int c = 0;
            for(j = 0;j < k;j++)
                if(s[j].val > val[i-1])
                    tmp[c++] = s[j];

            for(j = 0;j < c;j++){
                int L = Search(tmp[j].l, Hash, p);
                int R = Search(tmp[j].r, Hash, p)-1;
                //printf("%d %d\n", tmp[j].l, tmp[j].r);
                update(L, R, 1, tmp[j].c, 0, p-1);
                //printf("%I64d\n", sum[1]);
                ans += sum[1]*(val[i] - val[i-1])*(tmp[j+1].h - tmp[j].h);
                //printf("%I64d\n", sum[1]);
            }
        }
        printf("Case %d: %I64d\n", Case++, ans);
    }
}